Perform the indicated operations: \( \frac{2}{x-2}+\frac{x}{x+9}-\frac{x+20}{x^{2}+7 x-18} \)

To perform the indicated operations, we need to combine the given fractions. First, let's analyze the denominators: 1. The first term has a denominator of \( x - 2 \). 2. The second term has a denominator of \( x + 9 \). 3. The third term's denominator \( x^2 + 7x - 18 \) can be factored. Factoring \( x^2 + 7x - 18 \), we can find two numbers that multiply to -18 and add to 7, which are 9 and -2. Thus, \[ x^2 + 7x - 18 = (x - 2)(x + 9). \] Now, we can rewrite the expression: \[ \frac{2}{x - 2} + \frac{x}{x + 9} - \frac{x + 20}{(x - 2)(x + 9)}. \] Next, let's find the common denominator, which is \( (x - 2)(x + 9) \). We'll rewrite each term with this common denominator: 1. For \( \frac{2}{x - 2} \), we multiply the numerator and the denominator by \( x + 9 \): \[ \frac{2(x + 9)}{(x - 2)(x + 9)}. \] 2. For \( \frac{x}{x + 9} \), we multiply the numerator and the denominator by \( x - 2 \): \[ \frac{x(x - 2)}{(x - 2)(x + 9)}. \] 3. The third term is already using the common denominator: \[ -\frac{x + 20}{(x - 2)(x + 9)}. \] Now we can combine these fractions: \[ \frac{2(x + 9) + x(x - 2) - (x + 20)}{(x - 2)(x + 9)}. \] Expand each term in the numerator: - \(2(x + 9) = 2x + 18\), - \(x(x - 2) = x^2 - 2x\), - So, the numerator becomes: \[ 2x + 18 + (x^2 - 2x) - (x + 20). \] - This simplifies to: \[ x^2 + (2x - 2x - x) + (18 - 20) = x^2 - x - 2. \] Thus, we have: \[ \frac{x^2 - x - 2}{(x - 2)(x + 9)}. \] Next, we need to factor the numerator \(x^2 - x - 2\). We can find two numbers that multiply to -2 and add to -1, which are -2 and 1: \[ x^2 - x - 2 = (x - 2)(x + 1). \] Now substituting it back into our expression yields: \[ \frac{(x - 2)(x + 1)}{(x - 2)(x + 9)}. \] As long as \(x \neq 2\), we can cancel \(x - 2\): \[ \frac{x + 1}{x + 9}. \] Finally, the result is: \[ \frac{x + 1}{x + 9}, \quad x \neq 2. \]